Linear Antiferromagnetic Chain with Anisotropic Coupling

TLDR

The study discusses long‑range order in antiferromagnetic chains and critiques the variational method’s predictions. The authors solve exactly a linear spin‑½ chain with anisotropic coupling, varying the anisotropy parameter α between 0 and 1, and compute the ground‑state energy and short‑range order from its α‑dependence. They show that the kink in the short‑range order predicted by the variational method is spurious, eliminating the supposed discontinuity in the second derivative of the energy.

Abstract

The exact solution is given for a linear chain of $N$ atoms of spin \textonehalf{} coupled together by the anisotropic Hamiltonian $\mathcal{H}=2J\ensuremath{\Sigma}\stackrel{N}{i=1}[{{S}_{i}}^{z}{{S}_{i+1}}^{z}+(1\ensuremath{-}\ensuremath{\alpha})({{S}_{i}}^{x}{{S}_{i+1}}^{x}+{{S}_{i}}^{y}{{S}_{i+1}}^{y})].$ The energy of the antiferromagnetic ground state is computed and comparison is made with a variational method. The parameter $\ensuremath{\alpha}$ is allowed to vary between 0 and 1, regulating the relative amount of Ising anisotropy. The short-range order, $\ensuremath{\Sigma}{i}^{}{{S}_{i}}^{z}{{S}_{i+1}}^{z}$, is calculated exactly from the variation of the ground-state energy with $\ensuremath{\alpha}$. It is shown that a kink in the short-range order curve calculated using the variational method is fictitious, and the associated discontinuity in $\frac{{\ensuremath{\partial}}^{2}E}{\ensuremath{\partial}{\ensuremath{\alpha}}^{2}}$ is nonexistent. A discussion is given of long-range order and criticisms are presented regarding the predictions of the variational method.